So I came across a problem "How to reverse an integer in javascript?" I successfully managed to reverse the positive numbers for eg if I enter 123 then I get the output 321 but on the hand, if I am trying some negative number like -123 then I get 0 as the output. How can I solve this issue and get output as -321?

var reverse = function(x){
    let a = 0;
    while(x>0){
        a = a * 10   x; 
        // 0 = 0 *10   123 = 3
        // a=3
        // 0 = 0 *10   12 = 2
        // a=2
        // 0 = 0*10   1 = 1
        // a=1
        x = Math.floor(x/10)
         //  x = 123/10 = 12
        //  x = 12/10 =1
    }
    console.log(a);
    return a;
}
var x = -123;
reverse(x)

CodePudding user response：

Just Preserve the sign of an integer, like

var reverse = function(x){
    let sign = x<0?-1:1;
    let a = 0;
    x=Math.abs(x);
    while(x>0){
        a = a * 10   x; 
        // 0 = 0 *10   123 = 3
        // a=3
        // 0 = 0 *10   12 = 2
        // a=2
        // 0 = 0*10   1 = 1
        // a=1
        x = Math.floor(x/10)
         //  x = 123/10 = 12
        //  x = 12/10 =1
    }
    console.log(sign*a);
    return a;
}
var x = -123;
reverse(x)

CodePudding user response：

Convert negative integer to positive by multiplying with -1, then reverse then convert it back to negative

var reverse = function(x){
    let isNegative=x<0;
    x=Math.abs(x);
    let a = 0;
    while(x>0){
        a = a * 10   x; 
        // 0 = 0 *10   123 = 3
        // a=3
        // 0 = 0 *10   12 = 2
        // a=2
        // 0 = 0*10   1 = 1
        // a=1
        x = Math.floor(x/10)
         //  x = 123/10 = 12
        //  x = 12/10 =1
    }
    console.log(a);
    return isNegative ? a* -1 : a;
}

CodePudding user response：

let reverse = function(x){
    let a = 0;
    while(true){
        a = a * 10   x; 
        // 0 = 0 *10   123 = 3
        // a=3
        // 0 = 0 *10   12 = 2
        // a=2
        // 0 = 0*10   1 = 1
        // a=1
        x = x > 0 ? Math.floor(x/10) : Math.ceil(x/10);
        //  x = 123/10 = 12
        //  x = 12/10 =1
      if(x === 0){
        break;
      }
    }
    console.log(a);
    return a;
}
let x = -123;
reverse(x);

Explanation :

1. Changed while loop condition from x>0 to true to enable entering into loop when the argument is less than or equal to zero
2. Math.floor() round the number to the nearest smaller integer. Which means

Math.floor(10.5); Console output : 10

Math.floor(-10.5); Console output : -11

Incorporated the logic using Math.ceil() function when the number is less than zero.

CodePudding user response：

parse to string -> reverse -> parse to number back

function reverse(num){
  return num < 0 ? 
    ((num * -1) '').split('').reverse().join('') *-1 
    : (num '').split('').reverse().join('') *1
}

CodePudding user response：

var negativeNumber = -9813;

function convertToPositive(number) {
  if (number < 0) {
    var positiveNum = Math.abs(number);
    var stringNum = positiveNum.toString();
    var numArray = stringNum.split('');
    var reverseNum = numArray.reverse();
    var reverseString = reverseNum.join('');
    var reverseNum = parseInt(reverseString, 10);
    var negativeReverseNum = reverseNum * -1;
    return negativeReverseNum;
  }
}

convertToPositive(negativeNumber);